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प्रश्न
Draw a straight line AB of 9 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
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उत्तर
Steps of oonstruction:
(i) Draw a line segment AB of 9 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
Proof:
(i) Take any point on LM say P.
(ii) Join PA and PB.
Since, Plies on the right bisector of line AB.
Therefore, Pis equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.
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संबंधित प्रश्न
Use ruler and compasses only for this question:
I. Construct ABC, where AB = 3.5 cm, BC = 6 cm and ABC = 60o.
II. Construct the locus of points inside the triangle which are equidistant from BA and BC.
III. Construct the locus of points inside the triangle which are equidistant from B and C.
IV. Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and records the length of PB.
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